Question: $\begin{cases}a(1)=-2\\\\ a(n)=a(n-1)-5 \end{cases}$ Find the $4^{\text{th}}$ term in the sequence.
This is a recursive formula. It tells us that the first term is $-2$ and that the common difference is $-5$. $\begin{aligned} {a(1)}&=-2 \\\\ {a(2)}&={a(1)}-5=-7 \\\\ {a(3)}&={a(2)}-5=-12 \\\\ {a(4)}&={a(3)}-5=-17 \end{aligned}$ The $4^{\text{th}}$ term is $-17$.